Moduli of Right-Angled Hyperbolic Hexagons

Alternating side lengths as natural coordinates on moduli space.

Right angled hyperbolic hexagons have proven to play a foundational role in the understanding of surfaces: two can be glued to make a pair of pants, and all orientable surfaces can then be built from the resulting pants.

Computing $X,Y,Z$ in terms of $x,y,z$.

Here’s an explicit computation in hyperbolic trigonometry, starting from three side lengths $x,y,z$ and giving the unique lengths of the opposing three sides $X,Y,Z$ which produce a right angled hexagon. Because the situation is symmetric in the labels we shall just compute one of them, finding $Y$ as a function of $x,y$ and $z$.

To begin, we again draw the common perpendicular to the $Y,y$ sides, dividing the hexagon into the same pair of right angled pentagons as above. Both the edges $Y$ and $y$ have been subdivided, into segments of length $h,k$ and $u,v$ respectively, by this perpendicular of length $d$. Call the resulting pentagons $P$ and $Q$.

The quantity we want is $Y=h+k$, and since $Y$ is a length it is enough to pin down $\cosh Y$. The angle-sum law for the hyperbolic cosine turns that into a question about the two pieces separately:

$\cosh Y=\cosh(h+k)=\cosh h\,\cosh k+\sinh h\,\sinh k$

So we need the two products $\cosh h\,\cosh k$ and $\sinh h\,\sinh k$. Both pentagons hand these to us, through the single identity that governs a right angled pentagon: any two adjacent sides $A,B$ determine it completely, and the side $D$ two steps further round comes out as

$\cosh D=\sinh A\,\sinh B$

Everything below is this one fact, applied inside $P$ or inside $Q$.

Applying it in $P$ and then in $Q$, and allowing ourselves $X$ and $Z$ for the moment:

$$\begin{aligned} \cosh h &= \sinh Z\,\sinh v, &\qquad \cosh k &= \sinh X\,\sinh u,\\[4pt] \sinh h\,\sinh x &= \cosh v, &\qquad \sinh k\,\sinh z &= \cosh u. \end{aligned}$$

We do not know $X$ or $Z$. But the same relation, read across the cut, expresses both of their hyperbolic sines through the single unknown $d$:

$\sinh Z\,\sinh x=\cosh d=\sinh X\,\sinh z$

which is exactly what is needed to eliminate them. Substituting into the two products, and noticing that both land over the same denominator:

$$\begin{aligned} \cosh h\,\cosh k &= \sinh Z\,\sinh v\,\sinh X\,\sinh u\\[4pt] &= \frac{\cosh d\,\sinh v}{\sinh x}\cdot\frac{\cosh d\,\sinh u}{\sinh z} = \frac{\cosh^2 d\,\sinh u\,\sinh v}{\sinh x\,\sinh z},\\[10pt] \sinh h\,\sinh k &= \frac{\cosh v}{\sinh x}\cdot\frac{\cosh u}{\sinh z} = \frac{\cosh u\,\cosh v}{\sinh x\,\sinh z}. \end{aligned}$$

Adding them gives $\cosh Y$ outright:

$\cosh Y=\frac{\cosh^2 d\,\sinh u\,\sinh v+\cosh u\,\cosh v}{\sinh x\,\sinh z}$

The only quantity here that is not a side length is $d$, and it comes out by splitting $\cosh^2 d=\sinh^2 d+1$. The piece that splits off is precisely the angle-sum law again, this time for $y=u+v$:

$$\begin{aligned} \cosh Y\,\sinh x\,\sinh z &= \cosh^2 d\,\sinh u\,\sinh v+\cosh u\,\cosh v\\[4pt] &= \sinh^2 d\,\sinh u\,\sinh v+\bigl[\sinh u\,\sinh v+\cosh u\,\cosh v\bigr]\\[4pt] &= \sinh^2 d\,\sinh u\,\sinh v+\cosh y \end{aligned}$$

One term left, and the picture disposes of it. In $Q$ the sides $u$ and $d$ are adjacent, and in $P$ so are $v$ and $d$; in each case the pentagon relation returns the side opposite that pair, which is one of the lengths we started with.

$\sinh u\,\sinh d=\cosh z,\qquad\qquad \sinh v\,\sinh d=\cosh x$

Multiplying these together turns the stray term into $\cosh x\,\cosh z$, and the whole right-hand side is now written in $x,y,z$ alone:

$\cosh Y\,\sinh x\,\sinh z=\cosh x\,\cosh z+\cosh y$

Solving for $\cosh Y$:

$\cosh Y=\frac{\cosh x\,\cosh z+\cosh y}{\sinh x\,\sinh z}$

Nothing in the argument depended on which of the three opposite pairs we singled out, so permuting $(X,x),(Y,y),(Z,z)$ delivers all three formulas at once — the statement boxed at the top of the page.

The hexagon law of sines

Before drawing any conclusions, one more fact, proved by the same technique: the same cut, applied to a different pair of opposite sides.

We derive the equality just for the $x$ and $y$ sides; the other pairs are identical. Drop the common perpendicular between $z$ and $Z$ this time. It splits the hexagon into two right angled pentagons as before, and the side they now share — call it $L$ — can be reached from either one:

$\sinh X\,\sinh y=\cosh L=\sinh x\,\sinh Y$

Two expressions for one length, so dividing through gives the stated equality.

Right-angled hexagons as hyper-ideal triangles

The resemblance to the triangle law of sines is not a coincidence, and seeing why explains where every formula on this page comes from.

Extend the three sides $X,Y,Z$ to complete geodesics. Each pair is ultraparallel — they miss each other entirely — so together they form a triangle whose vertices have run off past the boundary: a hyper-ideal triangle. Every pair of ultraparallel geodesics has a unique common perpendicular, and adding those three in cuts our hexagon back out of it. So a right angled hexagon and a hyper-ideal triangle are the same object described twice.

What makes this useful is that hyperbolic trigonometry does not notice the transition. As two geodesics separate and stop meeting, the angle $\theta$ between them is replaced by the length $d$ of their common perpendicular, and the identities carry over under

$\cos\theta\ \longmapsto\ \cosh d,\qquad \sin\theta\ \longmapsto\ \pm\sinh d$

An ordinary triangle’s angles are our hexagon’s alternating sides $x,y,z$, and its sides are $X,Y,Z$. Under that dictionary the triangle law of sines reads

$$\frac{\sin\chi}{\sinh x}=\frac{\sin\upsilon}{\sinh y}=\frac{\sin\zeta}{\sinh z} \qquad\rightsquigarrow\qquad \frac{\sinh X}{\sinh x}=\frac{\sinh Y}{\sinh y}=\frac{\sinh Z}{\sinh z}$$

which is exactly the law of sines we just proved by hand.

The same dictionary carries the second hyperbolic law of cosines across, and that one is worth writing down: for a triangle it determines a side from the three angles,

$\cos\chi=-\cos\upsilon\,\cos\zeta+\sin\upsilon\,\sin\zeta\,\cosh X$

so for a hexagon it determines one side from the opposite triple of alternating sides,

$\cosh x=-\cosh y\,\cosh z+\sinh y\,\sinh z\,\cosh X$

and rearranging,

$\cosh X=\frac{\cosh y\,\cosh z+\cosh x}{\sinh y\,\sinh z}$

which is the theorem at the top of the page. So anyone already holding the trigonometry of hyper-ideal triangles can have the moduli of right angled hexagons in three lines.

We did not take that route, and deliberately: the computation above imports nothing, building everything from the trigonometry of right angled pentagons we already had in hand.